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I'm not very mathematical, so yea, sorry... I know that the ratio when dividing any two fibonacci numbers approaches the golden ratio. Is there any way to find what the ratio when dividing two generalised fibonacci sequences approaches?

I have two sequences, $h$ and $k$, where $$ h_{n}=(n+1)h_{n-1} + h_{n-2}\\ k_{n}=(n+1)k_{n-1} + k_{n-2} $$

$ h_{0}=0, h_{1} = 1\\ k_{0}=1, k_{1}=2 $

and I need $\lim\limits_{n \to +\infty} \frac{h_n}{k_n}$

So approximations start as $ \frac{1}{2}, \frac{3}{7}, \frac{13}{30} $ etc etc. I'm not sure if this helps, but when searching some of those approximations, one of them flagged up an integer series, the approximation $\frac{421}{972}$ showed me this: https://oeis.org/A058294 The following approximation in the sequence was also there, $\frac{3015}{6961}$

Alternatively, I know my problem could be solved if I could evaluate $$ \sum_{n=0}^{\infty} \frac{(-1)^n}{k_n \times k_{n+1}} $$ But I'm guessing the 1st would be much easier, just tell me if I'm wrong :) thanks.

Chris
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  • Do you know how to solve the recurrences for $h_n$ and for $k_n$? From those solutions it should be easy to work out the ratio. – Gerry Myerson Feb 09 '20 at 11:54
  • @GerryMyerson Hi, do you mean solving the recurrence relations to a different type of formula, that could be evaluated? If so I don't think it's possible, I tried a few sites that solve that type of thing, and they couldn't solve it. – Chris Feb 09 '20 at 11:55
  • $h(n)$ is given by http://oeis.org/A058307 (although I'm not sure it's helpful), and $k_n$ might be in the oeis, too. But maybe there's a better way. – Gerry Myerson Feb 09 '20 at 12:01
  • @GerryMyerson Not sure if that's useful, but that might explain why some of the numbers were popping up in other OEIS sequences. Thanks! I'll look into it if I can't find a better way. I'll look for $k_n$ too.

    There may be a better way, such as the summation I left at the bottom, but that seems equally as hard haha

     – Chris Feb 09 '20 at 12:04
  • @GerryMyerson Hi I found it, it's also there with an interesting comment! [link]https://oeis.org/A001053 It references the problem I'm trying to solve, a continued fraction. It says 'Denominator of continued fraction given by C(n) = [ 1; 2,3,4,...n ]' And I am trying to solve [0;2,3,4,5,6,...] – Chris Feb 09 '20 at 12:06
  • For the above comment, it turns out it's not exactly A001053, they use different starting values, so the sequence starts after the first 2 values. – Chris Feb 09 '20 at 12:31
  • The continued fraction looks a lot like a special case of Gauss's continued fraction based on $ _0F_1$. See https://en.wikipedia.org/wiki/Gauss%27s_continued_fraction. – random Feb 09 '20 at 16:27
  • @random Yea it does sort of, but it's been solved now :d I'm annoyed, since I saw the Bessel function come up on wolfram alpha, but I had no idea what it was, luckily this person managed to explain the result below to me :) – Chris Feb 09 '20 at 16:59
  • If you had told us from the start that it was that continued fraction, I would have referred you to a paper of D H Lehmer, from the 1930s I think, where he evaluated all continued fractions with partial quotients in arithmetic progression in terms of Bessel functions. In essence, your question is a duplicate of https://math.stackexchange.com/questions/342518/calculate-an-infinite-continued-fraction – Gerry Myerson Feb 09 '20 at 22:04
  • https://oeis.org/A016825/a016825.pdf is the Lehmer paper. – Gerry Myerson Feb 09 '20 at 22:08
  • Thanks, I'll have a read :) – Chris Feb 09 '20 at 22:24

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Okay, I've solved it thanks a legend in a discord server :) The limit is equivalent to

$$ \frac{I_2(2)}{I_1(2)} $$

Which itself is equivalent to

$$ \frac{2}{3} \times \frac{ \int_{0}^\pi e^{-2cos(t)} sin^4(t) dt }{ \int_{0}^\pi e^{-2cos(t)} sin^2(t) dt } $$

Where $I_n$ is a modified Bessel function.

Chris
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